Catalyst End

Colligative Properties Lab

Lecture 4.7 – Colligative Properties

Today’s Learning Targets LT. 4.11 – I can express the concentration of a solution in terms of mass percentage, mole fraction, molarity, and molality. LT 4.13 – I can explain the concept of colligative properties and calculate the freezing point depression and/or the boiling point elevation of a given solution.

Solute + Solvent  Solution Solution Chemistry Recall, Solute - The substance in the smaller amount. Substance being dissolved Solvent – The substance in the larger amount. Substance that does the dissolving. Solute + Solvent  Solution

Mass Percent We can determine the percent of a solute in a solution by using the equation:

Parts per Million (ppm) Oftentimes, the concentration of a solute is very low when compared to the solvent. We use parts per million in this case:

Class Example A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. What is the mass percentage of solute in this solution?

Table Talk A 2.5 g sample of groundwater was found to contain 5.4 μg of Zn2+. What is the concentration of Zn2+ in parts per million?

Molarity Recall, the molarity of the solution can be calculated by:

Molality We can also calculate the molality of a solution by:

Colligative Properties Colligative Properties are properties that depend only on the quantity of a substance, not the identity of the substance. The following are all colligative properties: Vapor Pressure Lowering Boiling – Point Elevation Freezing Point Depression Osmotic Pressure

Boiling Point Elevation When you dissolve a solute in a solution, you create more substances for the solvent to interact with. This increased interaction makes it harder for the solvent to escape, so the boiling point is increased. We can calculate this elevation using the equation: Tb = Increase in boiling point i = Vant hoff factor (how many ions formed when dissolved) Kb = Molal boiling point depression constant m = Molality of solution

Freezing Point Depression When you dissolve a solute in a solution, you create more substances for the solvent to interact with. This increased interaction makes it harder for the solvent to form the ordered solid arrangement. We can calculate this elevation using the equation: Tf = Change in freezing point i = Vant hoff factor (how many ions formed when dissolved) Kf = Molal freezing point depression constant m = Molality of solution

Class Example Automotive antifreeze consists of ethylene glycol, CH2(OH)CH2(OH), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 % solution of ethylene glycol in water

Table Talk Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus trees. The typical freezing point is CHCl3 is -63.5 oC.

Molar Mass from Colligative Properties From colligative properties, the molar mass of an unknown substance can be determined if mass of dissolved solute, mass of solvent, change in freezing/boiling point, and K are known.

Class Example A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution was 0.357 oC higher than that of the pure solvent. Calculate the molar mass of the solute. 88.0 g/mol

Table Talk Camphor (C10H16O) melts at 179.8 oC, and it has a particularly large freezing point depression constant, Kf = 40.0 oC/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 oC. What is the molar mass of the solute? 110 g/mol

White Board Problems

Question 1 Which of the following aqeuous solutions has the highest boiling point? a. 0.5 m NaCl b. 0.5 m KBr c. 0.5 m CaCl2 d. 0.5 m C6H12O6 e. 0.5 m NaNO3

Question 2 You dissolve 2 moles of NaCl in 450. kg of H2O. What is the freezing point depression if Kf for this solution is 1.86 oC/m?

Question 3 How many moles of Na2SO4 must be added to 500 mL of water to create a solution that has a 2 M concentration of Na+?

Question 4 When 31.0 grams of a nonionic substance is dissolved in 2.00 kg of water, the observed freezing – point depression of the solution is 0.93 oC. If Kf is 1.86 oC/m, then what is the molar mass of this substance?

Question 5 How many grams of MgSO4 are in 100.0 mL of a 5 M solution?

Closing Time Read 13.4 and 13.5 (533 to 535 and 539) and answer the essential questions There will be a formal lab report on this lab! Do book problems: